Chapter 6.5, Problem 37AYU

In Problems 17-40, graph each function. Be sure to label key points and show at least two cycles. Use the graph to determine the domain and the range of each function.

$y=\frac{1}{2}\tan \left(\frac{1}{4}x\right)-\text{2}$

To determine

To find: Domain and range.

Answer to Problem 37AYU

Solution:

Domain is $\left\{x\text{|}x\ne 2\pi k,k\text{ is an odd integer}\right\}$.

Range $=\{y│-\infty <y<\infty \}$.

Explanation of Solution

Given:

$y=\frac{1}{2}\tan \left(\frac{1}{4}x\right)-2$

Calculation:

The graph $y=\frac{1}{2}\tan \left(\frac{1}{4}x\right)-2$.

By the graph we determine that the domain of this function is the set of all real numbers where cosecant is defined, which excludes points $x$ where vertical asymptotes occur. Since $\tan \left(\frac{1}{4}x\right)=\frac{\sin \left(\frac{1}{4}x\right)}{\cos \left(\frac{1}{4}x\right)}$ this happens when $\cos \left(\frac{1}{4}x\right)=0\Rightarrow \left(\frac{1}{4}x\right)=\frac{\left(2k+1\right)\pi }{2}\Rightarrow x=2\left(2k+1\right)\pi$. Therefore the domain is $\left\{x\text{|}x\ne 2\left(2k+1\right)\pi ,k\text{ is an integer}\right\}$. Due to the positive and negative vertical asymptotes the range is $\{y│y\le │-\infty <y<\infty \}$. This agrees with the graph provided. (Notice the vertical shift doesn’t change the domain or the range).

Therefore the,

Domain is $\left\{x\text{|}x\ne 2\pi k,k\text{ is an odd integer}\right\}$.

Range $=\{y│-\infty <y<\infty \}$.